0

$$\left[\frac{2222^{5555}}{7} + \frac{5555^{2222}}{7}\right]$$

Please guide me through steps. Thanx..

user2345215
  • 16,422
Roshan
  • 123
  • 2
    Show your effort. – johny Apr 02 '14 at 12:23
  • At least the third time this question has come up. http://math.stackexchange.com/questions/519137/determine-the-remainder-of-dividing-2222555555552222-by-7 --- also related is http://math.stackexchange.com/questions/352607/prove-that-2222555555552222-3333555544442222-pmod-7 --- and http://math.stackexchange.com/questions/279333/what-will-be-the-ones-digit-of-the-remainder-in-left55552222-2222555 – Gerry Myerson Apr 02 '14 at 12:32

2 Answers2

0

We need to evaluate $$2222^{5555}+5555^{2222}\equiv?\pmod{7}$$

We have,

$$2222\equiv3\pmod{7}$$ $$3^5\equiv5\pmod{7}$$

$$5555\equiv4\pmod{7}$$

$$4^2\equiv2\equiv-5\pmod{7}$$

Can you do it now?

Guy
  • 8,857
  • 1
  • 28
  • 57
  • I think $5555\equiv4\pmod7$ is irrelevant. What you need is the reduction modulo 6, not modulo 7. – Gerry Myerson Apr 02 '14 at 12:30
  • 1
    @GerryMyerson I was thinking $$2222^{5555}+5555^{2222}\equiv 5^{1111}-5^{1111}\equiv0\pmod{7}$$. I think it is valid, unless I made a arithmetic mistake somewhere. – Guy Apr 02 '14 at 12:32
  • No, that's good. I retract my comment. – Gerry Myerson Apr 02 '14 at 12:36
  • @GerryMyerson ah okay. :) but if I may ask, how where you proposing to do it via modulo 6. Sounds interesting. I am sure I will learn something. – Guy Apr 02 '14 at 12:54
  • 1
    The idea is that if $m\equiv n\pmod6$, then $a^m\equiv a^n\pmod7$. So $5555\equiv5\pmod6$ implies $2222^{5555}\equiv3^5\pmod7$. – Gerry Myerson Apr 02 '14 at 22:34
  • @GerryMyerson ah wow. that's brilliant. Starting now I will able to use these "cycles" more efficiently. :) – Guy Apr 03 '14 at 03:50
0

$$a^{\varphi(n)} \equiv 1 \pmod n, \text{ when } (a,n)=1$$ $$\varphi(7)=6 \Rightarrow 2222^6\equiv 1 \pmod7$$ $$2222^{5555} \pmod7=2222^{925 \cdot 6 +5} \pmod7=(2222^6)^{925} \cdot 2222^5 \pmod7 = 2222^5 \pmod7 =3^5 \pmod7=243 \pmod7=5$$ Do the same for the other one..

Mary Star
  • 13,956